1963 AHSME Problems/Problem 35

Problem

The lengths of the sides of a triangle are integers, and its area is also an integer. One side is $21$ and the perimeter is $48$ . The shortest side is:

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$

Solution

Let $b$ and $c$ be the other two sides of the triangle. The perimeter of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.

By Heron's Formula, the area of the triangle is $\sqrt{24 \cdot 3(24-b)(b-3)}$ . Plug in the answer choices for $b$ and write the prime factorization of the product to make sure it is a perfect square.

Testing $b = 8$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 16 \cdot 5} = \sqrt{2^7 \cdot 3^2 \cdot 5}$ , so $8$ does not work. However, testing $b = 10$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 14 \cdot 7} = \sqrt{2^4 \cdot 3^2 \cdot 7^2}$ , so $10$ works. The third side is $17$ , and the sides satisfy the Triangle Inequality, so the answer is $\boxed{\textbf{(B)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 36
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1963 AHSME Problems/Problem 35

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